CodeForces - 1084A The Fair Nut and Elevator 数学

题目
The Fair Nut lives in n story house. ai people live on the i-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening.

It was decided that elevator, when it is not used, will stay on the x-th floor, but x hasn’t been chosen yet. When a person needs to get from floor a to floor b, elevator follows the simple algorithm:

Moves from the x-th floor (initially it stays on the x-th floor) to the a-th and takes the passenger.
Moves from the a-th floor to the b-th floor and lets out the passenger (if a equals b, elevator just opens and closes the doors, but still comes to the floor from the x-th floor).
Moves from the b-th floor back to the x-th.
The elevator never transposes more than one person and always goes back to the floor x before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the a-th floor to the b-th floor requires |a−b| units of electricity.
Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the x-th floor. Don’t forget than elevator initially stays on the x-th floor.

题意：
找一个最省电的电梯停留楼层。已知楼高和每层楼有多少人，每个人每天用电梯2次

题解：
模拟实际，耗电量=电梯去接人+电梯送到目的地+电梯回停留楼层
让电梯分别停留在1-n层，计算每次的耗电量用d数组保存，最后找出数组中的最小值即可

代码

#include<iostream>
#include<cmath>
using namespace std;
int d[107];               //x设在每层楼所用的电
int main()
{int n;cin>>n;               //n层楼int a[107];           //每层多少人for(int i=1;i<=n;i++) cin>>a[i];for(int j=1;j<=n;j++) //j为每次设定的x{int sum=0;for(int i=1;i<=n;i++)  //每层楼要耗得电数{sum+=(abs(j-i)+abs(i-1)+abs(j-1))*2*a[i];}d[j]=sum;}int t=100000000;for(int i=1;i<=n;i++){if(d[i]<t)t=d[i];}cout<<t<<endl;return 0;
}

感谢观看！

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